Calculus Homework
Assignment 4
1. Find all the local maxima, local minima, and saddle points of
1
the function f (x, y) = 2
.
x + 2y 2 − 1
[like §14.7 #27]
Sol. Solve
−2x
fx = − 2
=0
(x + 2y 2 − 1)2
−4y
=0
fy = − 2
(x + 2y 2 − 1)2
We have that (x, y) = (0, 0), which is the only critical point of
f . Since
fxx =
16xy
4(x2 − 6y 2 − 1)
6x2 − 4y 2 + 2
,
f
=
,
f
=
−
xy
yy
(x2 + 2y 2 − 1)3
(x2 + 2y 2 − 1)3
(x2 + 2y 2 − 1)3
So fxx (0, 0) = −2 < 0, fxy (0, 0) = 0, fyy (0, 0) = −4 and thus
2
fxx fyy − fxy
= 8 > 0 at (0, 0). Therefore f (x, y) has only one
local maximum at (0, 0) with value f (0, 0) = −1.
2. Find all the local maxima, local minima, and saddle points of
the function f (x, y) = x cos y.
[like §14.7 #29]
Sol. Solve
fx = cos y = 0
fy = −x sin y = 0
We have that (x, y) = 0, (2k+1)π
, k ∈ Z, which are all critical
2
points of f . Since
fxx = 0, fxy = − sin y, fyy = −x cos y
Thus we have fxx fyy −fxy 2 = − sin2 y, so fxx fyy −fxy 2 = −1 < 0
(2k+1)π
at each 0, (2k+1)π
,
k
∈
Z.
Therefore,
0,
are saddle
2
2
points of f for each k ∈ Z.
3. Find all the local maxima, local minima, and saddle points of
the function f (x, y) = 2 − 2x2 y 2 .
[like §14.7 #44(b)]
Sol. Solve
fx = −4xy 2 = 0
fy = −4x2 y = 0
This shows that all points lying on the x-axis or y-axis are
critical points of f . Since
fxx = −4y 2 , fxy = −8xy, fyy = −4x2
It is clear that the second derivative test fails in this case. Note
that f (x, y) = 2 − 2x2 y 2 ≤ 2. Hence f has local maxima 2. 4. Find the absolute maxima and minima of the function f (x, y) =
2x2 − 4x + y 2 − 4y + 1 on the rectangular plate 0 ≤ x ≤ 2, 0 ≤
y ≤ 3.
[like §14.7 #31]
Sol. Solve
fx = 4(x − 1) = 0
fy = 2(y − 2) = 0
we have that (x, y) = (1, 2), which is the only critical point of
f . Since
fxx = 4, fxy = 0, fyy = 2
So fxx fyy − fxy 2 = 8 > 0 and fxx > 0, so f has local minimum
at (1, 2) with value f (1, 2) = −5. Denote the given rectangular
region by D, that is,
D = {(x, y)0 ≤ x ≤ 2, 0 ≤ y ≤ 3}
which is shown as following
On OA, f (x, y) = f (0, y) = y 2 − 4y + 1 on 0 ≤ y ≤ 3. Then
f 0 (0, y) = 0 ⇒ y = 2. Since fyy = 2 > 0 and observe that
f (0, 0) = 1, f (0, 2) = −3, f (0, 3) = −2, so f has local minimum at (0, 2) with value −2 and local maximum at (0, 0) with
value 1.
On AB, f (x, y) = f (x, 3) = 2x2 − 4x − 2 on 0 ≤ x ≤ 2. Then
f 0 (x, 3) = 0 ⇒ x = 1. Since fxx = 4 > 0 and observe that
f (0, 3) = f (2, 3) = −2, f (1, 3) = −4, so f has local minimum
at (1, 3) with value −4 and local maximum at both (0, 3) and
(2, 3) with value −2.
On BC, f (x, y) = f (2, y) = y 2 − 4y − 1 = f (0, y) so f is local
minimum at (2, 2) with value −3 and local maximum at (2, 0)
with value 1.
On OC, f (x, y) = f (x, 0) = 2x2 − 4x + 1 on 0 ≤ x ≤ 2. Then
f 0 (x, 0) = 0 ⇒ x = 1. Since fxx = 4 > 0 and observe that
f (0, 0) = 1, f (1, 0) = −1, f (2, 0) = 1, so f has local minimum
at (1, 0) with value −1 and local maximum at both (0, 0) and
(2, 0) with value 1.
Therefore, the absolute maximum of f occur at (0, 0) and (2, 0)
with value 1, and the absolute minimum of f ’occur at (1, 2)
with value −5.
5. Find the points on the curve xy 2 = 2 nearest the origin.
[like §14.8 #5]
Sol. Note that to find the point on xy 2 = 2 nearest the origin is equivalent to find the point which minimize the function
f (x, y) = x2 +y 2 subject to the constraint g(x, y) = xy 2 −2 = 0.
Observe that
∇f = 2xi + 2yj
∇g = y 2 i + 2xyj
If ∇f = λ∇g, then we have following system of equations
2x = λy 2
2y = 2λxy
If y = 0, then x = 0. But (x, y) = (0, 0) is not on the curve
xy 2 = 2. This implies y 6= 0, and thus 1 = λx ⇒ x = λ1 ⇒ λ2 =
λy 2 ⇒ y 2 = λ22 . Since xy 2 = 2, so
2=
1 2
2
· 2 = 3 ⇒ λ3 = 1 ⇒ λ = 1
λ λ
λ
√
Hence we have that (x, y 2 )√= (1, 2) ⇒ (x, y) = (1, ± 2).
Therefore, the points (1, ± 2) are the points on the curve
xy 2 = 2 nearest the origin.
6. Find the dimensions of the rectangle of greatest area that can
2
2
be inscribed in the ellipse x9 + y16 = 1 with sides parallel to the
coordinate axes.
[like §14.8 #11]
Sol.
This is equivalent to find the point (x, y) with x, y > 0 which
maximize the function A(x, y) = (2x)(2y) = 4xy subject to
2
2
g(x, y) = x9 + y16 − 1 = 0. Observe that
∇A = 4yi + 4xj
y
2x
i+ j
∇g =
9
8
If ∇A = λ∇g, we have following system of equations
2λ
x
9
λ
4x =
y
8
4y =
λ
λ
Thus y = 18
x ⇒ 4x = λ8 18
x ⇒ x = 0 or λ = ±24 ⇒ x =
4
y = 0 or y = ± 3 x. But (0, 0) is not on the ellipse g(x, y) = 0.
Substituting y = ± 34 x into g(x, y) = 0, that is,
2
√
√
± 43 x
x2
3 2
+
=1⇒x=
⇒y=2 2
9
16
2
√
Hence the length
√ of the rectangle is 3 2 and the width of the
rectangle is 4 2.
7. Find the maximum and minimum values of f (x, y, z) = x+2y +
3z on the sphere x2 + y 2 + z 2 = 30.
[like §14.8 #23]
Sol. Let g(x, y, z) = x2 + y 2 + z 2 − 30. Observe that
∇f = i + 2j + 3k
∇g = 2xi + 2yj + 2zk
If ∇f = λ∇g, we have following system of equations
2λx = 1
2λy = 2
2λz = 3
1
then y =
Thus x = 2λ
g(x, y, z) = 0, we have
1
λ
= 2x, z =
3
2λ
= 3x. Substituting into
√
x2 + (2x)2 + (3x)2 − 30 = 14x2 − 30 = 0 ⇒ x = ±
√
√
105
7
Thus y = ± 2 7105 , z = ± 3 7105 . Since
√105 2√105 3√105 √
f ±
,±
,±
= ±2 105
7
7
7
Therefore,
the maximum value of f on x2 + y 2 + z 2 = 30 is
√
2 √
105 and the minimum value of f on x2 + y 2 + z 2 = 30 is
−2 105.
8. Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in a sphere of radius 3.
[like §14.8 #27]
Sol. This is equivalent to find the point (x, y, z) with x, y, z > 0
which maximize the function V (x, y, z) = (2x)(2y)(2z) = 8xyz
subject to g(x, y, z) = x2 + y 2 + z 2 − 9 = 0. Observe that
∇V = 8yzi + 8xzj + 8xyk
∇g = 2xi + 2yj + 2zk
If ∇V = λ∇g, we have following system of equations
8yz = 2λx
8xz = 2λy
8xy = 2λz
Thus λx2 = 4xyz = λy 2 ⇒ y 2 = x2 ⇒ y = ±x. Similarly,
z = ±x. Substituting into g(x, y, z) = 0, we have
√
x2 + (±x)2 + (±x)2 − 9 = 0 ⇒ x = 3
√
Thus y =√z = 3.√Therefore
√ the dimension of the rectangular
box are 2 3 by 2 3 by 2 3 for maximum volume.